\b[No.] Definability of a <function> means definability of its graph. The same additive <structure automorphism> $s(x)=-x$ would have to preserve the graph of multiplication. It sends the graph point $(1,1,1)$ to $(-1,-1,-1)$, which is not a multiplication-graph point because $(-1)(-1)=1\ne-1$. This contradicts <automorphism invariance of parameter-free definability>.

Solution

Past exam of the model theory course of the University of Oxford / 2011 / 1 / b / i / γ / Solution

 Solution (source code)